Inscribed square

Geometry Level 3

We inscribed the red square into a bigger square, as it is shown in the figure below. The lengths of the bigger square are 10 10 .

What is the area of the smaller, red square?

60 None of the others 50 70 40

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Áron Bán-Szabó by Áron Bán-Szabó , 17, Hungary

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1 solution

The area of each of the 4 peripheral triangles is 1 2 ( 10 sin ⁡ ( 1 5 ∘ ) ) ( 10 cos ⁡ ( 1 5 ∘ ) ) = 50 sin ⁡ ( 1 5 ∘ ) cos ⁡ ( 1 5 ∘ ) = 25 sin ⁡ ( 3 0 ∘ ) = 12.5 \dfrac{1}{2}(10\sin(15^{\circ}))(10\cos(15^{\circ})) = 50\sin(15^{\circ})\cos(15^{\circ}) = 25\sin(30^{\circ}) = 12.5 ,

where the identity sin ⁡ ( 2 x ) = 2 sin ⁡ ( x ) cos ⁡ ( x ) \sin(2x) = 2\sin(x)\cos(x) was used. Thus the sum of the areas of the 4 peripheral triangles is 4 × 12.5 = 50 4 \times 12.5 = 50 , and so the area of the red square is 1 0 2 − 50 = 50 10^{2} - 50 = \boxed{50} .

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